Holy Dielectric

1. The problem statement, all variables and given/known data
A large block of dielectric contains small cavities of various shapes that may be assumed not to disturb appreciably the polarization. Show that, inside a needle-like cavity parallel to P, E is the same as in the dielectric

2. Relevant equations

[tex]\oint_{S}\vec{E}\cdot \vec{da}=\frac{Q_{enc}}{\epsilon_0}[/tex]

3. The attempt at a solution
Not really sure how to attack this problem....

I can create a Gaussian object inside the dielectic and using Gauss's law....

Taking the Gaussian surface to be cylindrical, i get that [tex]E=\frac{\pi p_{b}}{2\epsilon_0}[/tex], where [tex]p_{b}[/tex] is the bound charge density inside the dielectric.

How would I calculate E inside of the cavity? Same method? Doesn't [tex]p_{b}=0[/tex]?

What makes you think the electric field is cylindrically symmetric? Is the dielectric linear?

Reading your problem statement, I would assume that the polarization [itex]\textbf{P}[/itex] inside the dielectric is approximately uniform, and that the dielectric need not be linear (which means [itex]\textbf{E}[/itex] and [itex]\textbf{D}[/itex] need not be uniform).

I think you'll just want to calculate the electric field due to a uniformly polarized needle with polarization [itex]-\textbf{P}[/itex] and then use thew superposition principle.

But for a skinny needle, how much surface area will there be at the ends....will the resulting surface bound charge be large or small?...If the needle is long these charges will be far apart....doesn't that reduce the field even more?

No you don't.In fact, you should be able to argue that E is much much larger inside the dielectric, than the field of the needle---which is why you can neglect the field of the needle entirely. You only need to use the superposition principle (superimpose a needle of negative polarization onto the slab to create the needle shaped cavity);

Bound surface charge is also zero because it only exists at the surface and I'm looking at a gaussian surface inside?

No, the bound surface charge is very large, and doesn't necessarily create a symmetric field; so just because the net flux is zero inside, doesn't mean the field is zero. (Gauss' Law requires symmetry to be useful!)

The only way I can see this working is if [tex]\textbf{E}_{\text{cavity}}=\textbf{E}_{\text{slab} }+\textbf{E}_{\text{needle}}[/tex] and [tex]\textbf{E}_{\text{needle}}[/tex] is neglible (say zero), then [tex]\textbf{E}_{\text{slab}}[/tex] is also zero. Therefore, [tex]\textbf{E}_{\text{cavity}}[/tex] is zero.

Giving end result [tex]\textbf{E}_{\text{cavity}}=0=\textbf{E}_{\text{needle}}[/tex]

You need to understand that [itex]\textbf{E}_{\text{cavity}}[/itex] represents the total field inside the cavity; [itex]\textbf{E}_{\text{slab}}[/itex] represents the field inside the cavity, due to just the dielectric slab; and [itex]\textbf{E}_{\text{needle}}[/itex] represents the field inside the cavity due to just a uniformly polarized needle (w/ polarization [itex]-\textbf{P}[/itex])....